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Homework Help: Area of one loop of the rose r=cos(3theta)

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Use double integrals to find the Area of one loop of the rose r=cos(3theta)

    I know how to solve this, the only question I have is why theta is between -pi/6 and pi/6. I don't understand where those two values come from.
  2. jcsd
  3. Jun 22, 2011 #2


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    With theta equal to -pi/6, 3theta= -pi/2 and r= cos(3theta)= cos(-pi/2)= 0. Similarly, if theta is pi/6, 3theta= pi/2 and r= cos(3theta)= cos(pi/2)= 0. The only point with r= 0 is the origin, no matter what theta is. That means that starting at -pi/6 and going to pi/6 you have closed a loop. starting at the origin and coming back to it. Since cosine in not 0 at any point between -pi/2 and pi/2, cos(3theta) is not 0 at any point between -pi/6 and pi/6 so that encloses only one loop which is what you want.,
  4. Jun 22, 2011 #3
    finally! Now i understand it! Thank you so much! :)
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